Theory_3_Solution

# Theory_3_Solution - 39th International Physics Olympiad Hanoi Vietnam 2008 Theoretical Problem No 3 Solution Solution 1 For an altitude change dz

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39th International Physics Olympiad - Hanoi - Vietnam - 2008 Theoretical Problem No. 3 / Solution Solution 1. For an altitude change , the atmospheric pressure change is : dz dp gdz ρ = − (1) where is the acceleration of gravity, considered constant, g is the specific mass of air, which is considered as an ideal gas: mp VR T μ == Put this expression in (1) : dp g dz pR T =− 1.1. If the air temperature is uniform and equals , then 0 T 0 dp g dz T After integration, we have : () () 0 0 e g z RT pz p = (2) 1.2. If ( ) ( ) 0 Tz T z Λ (3) then () 0 dp g dz p RT z ⎡⎤ −Λ ⎣⎦ (4) 1.2.1. Knowing that : ( ) 0 11 0 00 ln dT z dz Tz Λ Λ ∫∫ by integrating both members of (4), we obtain : ( ) ( ) 0 1 ln ln ln T z gg T R T μμ ⎛⎞ Λ ΛΛ ⎝⎠ 0 z 01 0 g R z p T Λ Λ (5) 1

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39th International Physics Olympiad - Hanoi - Vietnam - 2008 Theoretical Problem No. 3 / Solution 1.2.2. The free convection occurs if: ( ) () 1 0 z ρ > The ratio of specific masses can be expressed as follows: 1 0 1 00 0 g R zp z T z pT z T μ Λ ⎛⎞ Λ == ⎜⎟ ⎝⎠ The last term is larger than unity if its exponent is negative: 10 g R −< Λ Then : 0 029 9 81 K 0034 831 m .. . . g R × Λ> = = 2. In vertical motion, the pressure of the parcel always equals that of the surrounding air, the latter depends on the altitude. The parcel temperature parcel T depends on the pressure. 2.1. We can write: parcel parcel dT dT dp dz dp dz = p is simultaneously the pressure of air in the parcel and that of the surrounding air. Expression for parcel dT dp By using the equation for adiabatic processes and equation of state, we can deduce the equation giving the change of pressure and temperature in a quasi-equilibrium adiabatic process of an air parcel: const pV γ = 1 parcel const Tp = (6) 2
39th International Physics Olympiad - Hanoi - Vietnam - 2008 Theoretical Problem No. 3 / Solution where p V c c γ = is the ratio of isobaric and isochoric thermal capacities of air. By logarithmic differentiation of the two members of (6), we have: parcel parcel 1 0 dT dp Tp += Or parcel parcel 1 dT T dp p = (7) Note: we can use the first law of thermodynamic to calculate the heat received by the parcel in an elementary process: parcel V m dQ c dT pdV μ =+ , this heat equals zero in an adiabatic process. Furthermore, using the equation of state for air in the parcel parcel m pV RT = we can derive (6) Expression for dp dz From (1) we can deduce: dp pg g dz RT ρ =− where is the temperature of the surrounding air. T On the basis of these two expressions, we derive the expression for : parcel / dT dz parcel parcel 1 dT T g G dz R T γμ (8) In general, is not a constant.

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## This note was uploaded on 11/08/2011 for the course PHYS 0000 taught by Professor Na during the Spring '11 term at Rensselaer Polytechnic Institute.

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Theory_3_Solution - 39th International Physics Olympiad Hanoi Vietnam 2008 Theoretical Problem No 3 Solution Solution 1 For an altitude change dz

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